Two bodies of masses m and M are placed at a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. Then:
$V = - \frac{G}{d} (m + M)$
$V = \frac{G}{d}$
$V = - \frac{G}{d}$
$V = - \frac{G}{d} {(\sqrt{m} + \sqrt{M})}^{2}$

Question

Two bodies of masses m and M are placed at a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. Then:
$V = - \frac{G}{d} (m + M)$
$V = \frac{G}{d}$
$V = - \frac{G}{d}$
$V = - \frac{G}{d} {(\sqrt{m} + \sqrt{M})}^{2}$

Toppr Admin · Accepted Answer

Let that point be x distance from mThen -x2212-GM-d-x2212-x-2-x2212-Gmx2-x21D2-d-x2212-xx-x221A-MmPotential-x2212-GMd-x2212-x-x2212-GmxPotential-x2212-Gd-x221A-M-x221A-m-2

Two bodies of masses m and M are placed at a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. Then:V=−Gd(m+M)V=GdV=−GdV=−Gd(√m+√M)2

Let that point be x distance from mThen −GM(d−x)2=−Gmx2⇒d−xx=√MmPotential=−GMd−x−GmxPotential=−Gd(√M+√m)2

Two bodies of masses m and M are placed at a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V. Then:
$V = - \frac{G}{d} (m + M)$
$V = \frac{G}{d}$
$V = - \frac{G}{d}$
$V = - \frac{G}{d} {(\sqrt{m} + \sqrt{M})}^{2}$

Let that point be $x$ distance from $m$
Then $\frac{- G M}{(d - x)^{2}} = \frac{- G m}{x^{2}}$
$\Rightarrow \frac{d - x}{x} = \sqrt{\frac{M}{m}}$
$P o t e n t i a l = - \frac{G M}{d - x} - \frac{G m}{x}$
$P o t e n t i a l = \frac{- G}{d} (\sqrt{M} + \sqrt{m})^{2}$